Question

When a potential difference V is applied across a wire of resistance R, it dissipates energy at a rate W. If the wire is cut into two halves and these halves are connected mutually parallel across the same supply, the same supply, the energy dissipation rate will become:

• A. $\frac{1}{4}W$
• B. $\frac{1}{2}W$
• C. 4W
• D. 2W

Answer 1

The Correct Option is C

To solve this problem, we need to analyze the energy dissipation rate before and after the wire is cut.

Initial Energy Dissipation Rate: The power P dissipated by a resistor is given by:

\( P = \frac{V^2}{R} \),

where: - V is the voltage across the resistor, and R is the resistance.

Initially, with resistance R, the energy dissipation rate is:

\( W = P = \frac{V^2}{R} \).

After Cutting the Wire: When the wire is cut into two halves, each half has a resistance of:

\( R' = \frac{R}{2} \).

Connecting in Parallel: When these two halves are connected in parallel, the equivalent resistance R_eq is given by:

\( \frac{1}{R_{eq}} = \frac{1}{R'} + \frac{1}{R'} = \frac{2}{R/2} = \frac{4}{R} \).

Thus, the equivalent resistance is:

\( R_{eq} = \frac{R}{4} \).

New Energy Dissipation Rate: The new power P' dissipated in the circuit with the new resistance R_eq is:

\( P' = \frac{V^2}{R_{eq}} = \frac{V^2}{R/4} = \frac{4V^2}{R} \).

Comparing Power Dissipation Rates: Since \( W = \frac{V^2}{R} \) from the original circuit, we can relate the new power:

\( P' = 4W \).

Answer 2

The problem describes a scenario where a wire dissipates energy at a rate \(W\) when connected to a potential difference \(V\). We need to determine the new rate of energy dissipation when the wire is cut into two equal halves and these halves are connected in parallel to the same voltage supply.

Concept Used:

The solution relies on the fundamental principles of electrical resistance and power dissipation.

1. Electrical Power: The rate of energy dissipation (power, \(P\)) in a resistor is given by the formula: \[ P = \frac{V^2}{R} \] where \(V\) is the potential difference across the resistor and \(R\) is its resistance. This form is most useful here because the voltage supply \(V\) remains constant.
2. Resistance and Length: The resistance of a uniform wire is directly proportional to its length (\(L\)). If a wire of resistance \(R\) is cut into two equal halves, the resistance of each half will be \(R/2\).
3. Resistors in Parallel: When two resistors, \(R_1\) and \(R_2\), are connected in parallel, their equivalent resistance (\(R_{\text{eq}}\)) is given by: \[ \frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} \]

Step-by-Step Solution:

Step 1: Define the initial state.

In the initial setup, the full wire has a resistance \(R\). When a potential difference \(V\) is applied across it, the energy dissipation rate is \(W\).

Using the power formula, we can write:

\[ W = \frac{V^2}{R} \]

This equation serves as our reference for the initial power dissipation.

Step 2: Determine the resistance of the cut pieces.

The wire is cut into two equal halves. Since resistance is directly proportional to length, the resistance of each half (\(R_{\text{half}}\)) will be half of the original resistance \(R\).

\[ R_{\text{half}} = \frac{R}{2} \]

Step 3: Calculate the equivalent resistance of the parallel combination.

These two halves, each with resistance \(R_{\text{half}} = R/2\), are connected in parallel. Let the new equivalent resistance be \(R_{\text{parallel}}\).

\[ \frac{1}{R_{\text{parallel}}} = \frac{1}{R_{\text{half}}} + \frac{1}{R_{\text{half}}} = \frac{1}{R/2} + \frac{1}{R/2} \] \[ \frac{1}{R_{\text{parallel}}} = \frac{2}{R} + \frac{2}{R} = \frac{4}{R} \]

Solving for \(R_{\text{parallel}}\), we get:

\[ R_{\text{parallel}} = \frac{R}{4} \]

Step 4: Calculate the new energy dissipation rate.

The parallel combination is connected across the same voltage supply \(V\). Let the new energy dissipation rate be \(W_{\text{new}}\).

Using the power formula with the new equivalent resistance:

\[ W_{\text{new}} = \frac{V^2}{R_{\text{parallel}}} \]

Substitute the value of \(R_{\text{parallel}} = R/4\) into this equation:

\[ W_{\text{new}} = \frac{V^2}{(R/4)} \]

Final Computation & Result:

Simplifying the expression for the new power dissipation:

\[ W_{\text{new}} = 4 \left( \frac{V^2}{R} \right) \]

From Step 1, we know that the original power dissipation is \(W = \frac{V^2}{R}\). Substituting this into our expression for \(W_{\text{new}}\):

\[ W_{\text{new}} = 4W \]

Therefore, the energy dissipation rate will become 4W.